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Proof step 1 from proof same-used-extract (conclusion)
n < # wlar1 ⊦ same-used(wlar0, wlar1) ∧ wlar0[n] = wl-entry(m0, used) → wlar1[n].status = used
System applied rule normalize
And obtained the following new goals
Proof step 2
:
wlar0[n] = wl-entry(m0, used), n < # wlar1, same-used(wlar0, wlar1) ⊦ wlar1[n].status = used
Rule argument
Additional information